$\mathbb{Z}_n$ is a $\mathbb{Z}_{n^2}$ projective module? I try to apply the definition...
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1Does the obvious surjective homomorphism $p:\Bbb{Z}_{n^2}\to\Bbb{Z}_n$ split? – Jyrki Lahtonen Jun 12 '17 at 07:08
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@JyrkiLahtonen Actually that's my problem. How to prove that? – Problemsolving Jun 12 '17 at 07:22
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1Well, is there a morphism which splits that surjection? Have you tried to see what happens when $n$ is $2$, for example? – Mariano Suárez-Álvarez Jun 12 '17 at 07:39
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If $s:\Bbb{Z}n\to\Bbb{Z}{n^2}$ is a splitting homomorphism, what can you say about $s(1)$? A) for $s$ to be a homomorphism? B) for $p\circ s$ to be the identity? – Jyrki Lahtonen Jun 12 '17 at 08:24
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@JyrkiLahtonen Actually $s(1) = nk$ – Problemsolving Jun 12 '17 at 09:12
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Correct. And why is that a contradiction? – Jyrki Lahtonen Jun 12 '17 at 09:14
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How can $s$ be defined? Generators must go to generators. But this cannot defined here? – Jun 12 '17 at 09:15
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@JyrkiLahtonen Now $s(n)$ goes to 0, but it must go to 1?No? – Problemsolving Jun 12 '17 at 09:18
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@PregatireMatematica what they try to point out, is that the only such a morphism is the trivial one. – Jun 12 '17 at 09:32
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Correct! ${}{}$ – Jyrki Lahtonen Jun 12 '17 at 09:33
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The group $\mathbb{Z}/n\mathbb{Z}$ can be considered as a module over $\mathbb{Z}/n^2\mathbb{Z}$ by identifying it with the subgroup $n\mathbb{Z}/n^2\mathbb{Z}$.
The map $\mathbb{Z}/n^2\mathbb{Z}\to n\mathbb{Z}/n^2\mathbb{Z}$ defined by $x+n^2\mathbb{Z}\mapsto nx+n^2\mathbb{Z}$ is a surjective homomorphism. If the module is projective, then this homomorphism splits; however, $\mathbb{Z}/n^2\mathbb{Z}$ has a unique subgroup isomorphic to $n\mathbb{Z}/n^2\mathbb{Z}$, which is not a direct summand.
egreg
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The morphism $\mathbb Z_n \to \mathbb Z_{n^2}$ sent $1$ to $kn$. This can solve the problem
Jian
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@PregatireMatematica you need research the morphism,the composition can not be identity. What I have said is correct,but you give........... – Jian Jun 12 '17 at 09:50